A) -9 and -5
B) -5 and 9
C) 0 and 9
D) 5 and 9
Correct Answer: D
Solution :
Let \[y=\frac{{{x}^{2}}+34x-71}{{{x}^{2}}+2x-7}\] Þ \[{{x}^{2}}(y-1)+2(y-17)x+(71-7y)=0\] For real values of x, its discriminant \[D\ge 0\] \[\Rightarrow 4{{(y-17)}^{2}}-4(y-1)(71-7y)\ge 0\] \[\Rightarrow ({{y}^{2}}-3+y+289)-(71y-7{{y}^{2}}-71+7y)\ge 0\] \[\Rightarrow {{y}^{2}}-14y+45\ge 0\Rightarrow (y-5)(y-9)\ge 0\] It is possible when both \[y-5\] and\[y-9\] are negative or both positive. Let \[y-5\le 0\Rightarrow y\le 5\] and \[y-9\le 0\Rightarrow y\le 9\]. Hence \[y\le 5\] .....(i) If \[y-5\ge 0\Rightarrow \]\[y\ge 5\]and \[y-9\ge 0\Rightarrow y\ge 9\] Hence \[y\ge 9\]. .....(ii) Therefore y does not lie between 5 and 9.
You need to login to perform this action.
You will be redirected in
3 sec
